Yes, that is correct.
The "standard" quadratic formula is
[tex]\frac{-b\pm\sqrt{b^2- 4ac}}{2a}[/tex]
Of course, you can separate that into two fractions:
[tex]\frac{-b}{2a}\pm\frac{\sqrt{b^2- 4ac}}{2}[/tex]
If you take that second denominator, 2, inside the square root, it becomes 4:
[tex]\frac{-b}{2a}\pm\sqrt{\frac{b^2- 4ac}{4}}[/tex]
[tex]= \frac{-b}{2a}\pm\sqrt{\frac{b^}{4}- ac}[/tex]

Yes, that is correct.
The "standard" quadratic formula is
[tex]\frac{-b\pm\sqrt{b^2- 4ac}}{2a}[/tex]
Of course, you can separate that into two fractions:
[tex]\frac{-b}{2a}\pm\frac{\sqrt{b^2- 4ac}}{2}[/tex]
If you take that second denominator, 2, inside the square root, it becomes 4:
[tex]\frac{-b}{2a}\pm\sqrt{\frac{b^2- 4ac}{4}}[/tex]
[tex]= \frac{-b}{2a}\pm\sqrt{\frac{b^}{4}- ac}[/tex]

Presumably, he did that in order to simplify the arithmetic!

Er, Halls, what happened to the 'a' in the second denominator (I'm looking at at your work immediately following "Of course...")

I'm on a little medication after oral surgery, so if I missed something obvious I will apologize and blame that for my problem.